Respuesta :
Answer:0
Explanation:
Given
two particles of mass m and 3m
They move towards each other with velocity v
As the collision is Elastic therefore value of coefficient of restitution is 1
[tex]e=1[/tex]
[tex]e=\frac{Relative\ speed\ after\ collision}{Relative\ speed\ before\ collision}[/tex]
if [tex]u_1[/tex] and [tex]u_2[/tex] are velocity of approach of m and 3 m
and [tex]v_1[/tex] and [tex]v_2[/tex] are velocity after collision respectively
therefore
[tex]v_2-v-1=u_1-u_2[/tex]
[tex]v_2-v_1=v-(-v)[/tex]
[tex]v_2-v_1=2 v[/tex]
conserving momentum
[tex]m\cdot v+3m\cdot (-v)=3m\cdot (v_2)+m\cdot v_1[/tex]
[tex]-2mv=3mv_2+m(v_2-2v)[/tex]
[tex]4mv_2=0[/tex]
[tex]v_2=0[/tex]
Therefore velocity of heavier mass is zero
Answer:
Zero
Explanation:
m1 = m
m2 = 3m
u1 = v
u2 = - v
Let the final speed of m1 and m2 be v1 and v2 respectively.
Use conservation of momentum
m1 x u1 + 2 x u2 = m1 x v1 + m2 x v2
m x v - 3 m v = m v1 + 3m v2
- 2 mv = m (v1 + 3 v2)
v1 + 3 v2 = - 2v ... (1)
As the collision is perfectly elastic so the coefficient of restitution is 1.
Use Newton's formula
[tex]e = \frac{v_{2}-v_{1}}{u_{1}- u_{2}}[/tex]
v - (- v) = v2 - v1
v2 - v1 = 2 v ... (2)
Add equation (1) and (2), we get
v2 = 0
v1 = - 2v
So, the heavier body comes into rest after collision.
